Simplify; express your answer in exponential form. Assume $y\neq 0, q\neq 0$. $\dfrac{{(y^{-4})^{-4}}}{{(y^{5}q^{-2})^{3}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{-4}}$ to the exponent ${-4}$ . Now ${-4 \times -4 = 16}$ , so ${(y^{-4})^{-4} = y^{16}}$ In the denominator, we can use the distributive property of exponents. ${(y^{5}q^{-2})^{3} = (y^{5})^{3}(q^{-2})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{-4})^{-4}}}{{(y^{5}q^{-2})^{3}}} = \dfrac{{y^{16}}}{{y^{15}q^{-6}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{16}}}{{y^{15}q^{-6}}} = \dfrac{{y^{16}}}{{y^{15}}} \cdot \dfrac{{1}}{{q^{-6}}} = y^{{16} - {15}} \cdot q^{- {(-6)}} = yq^{6}$.